Preparation for Inverse Exponential Function

The exponential function is the function e^x, where e is the number (approximately 2.718281828) such that the function e^x equals its own derivative. The exponential function is used to model phenomena when a constant change in the independent variable gives the same proportional change (increase or decrease) in the dependent variable. (Source from Wikipedia).It is otherwise Written as exp(x).Here we are going to see about preparation about inverse exponential function and its example problems.

Preparation for finding the inverse of the function:

• First we got to to replace f(x) with y
• Switch x‘s   y‘s
• Solve for y
• Replace y of f ^–(x)                   

Preparation problems for preparation for inverse exponential function:

       pro1 :  Find the inverse function of the following y = 3x+1

Sol :  The given function is y = 3x +1

Following the above procdure to find the inverse of the function

First we have to Interchange x and y value

x=3y+1

Then now solve for y

Add both sides -1 then the equation will be

x-1 =3y + 1 -1

x-1=3y

Divide both sides 3 we get ,

y = `(x-1) / (3)`

Therefore the inverse function is  (x-1) / (3)

Example for inverse of exponential function:

        Pro 2:  Find the inverse of the following exponential function,
                    y = 5. 6 ^(3x-4) +7

Sol :   Interchange the x and y value then the equation will be,

x= 5. 6 ^(3y-4) +7

Now we solve for y

Add both sides -7 we get

x-7 = 5. 6 ^(3y-4) +7-7

x-7 = 5. 6 ^(3y-4)

Divide both sides 5 we get,

`(x-7) / (5)` = `(5. 6 (3y-4)) / (5)`

`(x-7) / (5)` = 6 ^(3y-4)

Take both sides log then the equation will be ,

Log ₆ `((x-7) / (5))` = log ₆ 6 ^(3y-4)

Log ₆ `((x-7) / (5))` = 3y-4

Add both sides +4

Log ₆ `((x-7) / (5))` + 4 = 3y

Both sides divide 3 we get

`{Log 6 ((x-7) / (5) + (4))} / (3)` = `(3y) / (3)`

`{Log 6 ((x-7) / (5) + (4))} / (3)` = y

Therefore the inverse exponential function is `{Log 6 ((x-7) / (5) + (4))} / (3)` = y

Hence the preparation of inverse exponential function is explained,

Figure Segment of a Circle

In geometry, a circular segment also circle segment. It is an area of a circle informally defined as an area which is "cut off" from the rest of the circle by a secant or a chord. The circle segment constitutes the part between the secant and an arc, excluding the circle's center.It is shown in figure bellow.
  

Understanding Circle Formulas is always challenging for me but thanks to all math help websites to help me out. 

Example of figure segment of a circle:

Example 1
Radius of the circle is 15 cm, angle of the Segment of a circle is 45 degree. Find the area of the Segment of a circle is shown in figure.


Solution:
Area of the circle = `1/2` r²`theta` - sin `theta`

                            = `1/2` r²( `pi/180` *`theta` - sin `theta`)                       

                             =0.5*15²(`pi/180` *45-(sin 45))

                             =112.5(0.785-0.7071)

   Area of the Segment of a circle=8.77 cm²

Example 2
          Radius of the circle is 10 cm, angle of the Segment of a circle is 55 degree. Find the area of the Segment of a circle is shown in figure.
 
Solution:

Area of the circle = `1/2` r²`theta` - sin `theta`

                            = `1/2` r²( `pi/180` *`theta` - sin `theta`)                       

                             =0.5*10²(`pi/180` *50-(sin 50))

                             =50(0.873-0.766)

   Area of the Segment of a circle=5.53 cm²

Example 3
Radius of the circle is 15 cm, angle of the Segment of a circle is 85 degree. Find the area of the Segment of a circle is shown in figure.

Solution:
Area of the circle = `1/2` r²`theta` - sin `theta`

                            = `1/2` r²( `pi/180` *`theta` - sin `theta`)                       

                             =0.5*15²(`pi/180` *85-(sin 85))

                             =112.5(1.484-0.996)

   Area of the Segment of a circle=54.9 cm²

Example 4
Radius of the circle is 15 cm, angle of the Segment of a circle is 105 degree. Find the area of the Segment of a circle is shown in figure.
Solution:
Area of the circle = `1/2` r²`theta` - sin `theta`

                            = `1/2` r²( `pi/180` *`theta` - sin `theta`)                       

                             =0.5*15²(`pi/180` *105-(sin 105))

                             =112.5(1.833-0.9659)

   Area of the Segment of a circle= 97.54 cm²

Example 5
Radius of the circle is 15 cm, angle of the Segment of a circle is 110 degree. Find the area of the Segment of a circle is shown in figure.

Solution:
Area of the circle = `1/2` r²`theta` - sin `theta`

                            = `1/2` r²( `pi/180` *`theta` - sin `theta`)                       

                             =0.5*15²(`pi/180` *110-(sin 110))

                             =112.5(1.92-0.9396)

Area of the Segment of a circle=110.29 cm²

Distribution of the Sample Means

A sample mean is a numerical set which is an average value of a particular portion of a number in a certain group. A sample mean is expressed as x.suppose if we take a sample of size n and there is a n independent variables x1,x2…xn and each value is respect to one randomly selected observation, then the distribution of the population  of these variables has the mean value mhu and the standard deviation sigma. Then the sample value is,  
                             X=1/n(x₁+x₂+…..x_n)

Mean and variance of the distribution of the sample means:

Using the mean and variance property of the random variable ,the mean and variance of the sample mean is given below.
µ_x= µ
sigmax=sigma/sqrt(n)`
therefore the mean value of the sample mean is similar to the mean of the population distribution and the variance is smaller than the variance of the population distribution.

For example:
In population distribution  where the mean value µ=20 and the standard deviation σ=2 that means ( N(20,2).then we have the simple random sample of 100 students ,then what is the  mean and variance of  the sample mean distribution?
                                                µ_x= µ
                                                µ_x=20
                                                σ_x=σ/`sqrt(n)`
                                                σ_x =2/`sqrt(100)`
                                                σ_x =2/10
                                                 σ_x =0.2

shape property of the distribution of the sample means:

It may be in the following shapes like normal, skewed, bimodal.

Explanation for distribution of the sample means:
If the population distribution is normal,then the distribution of  the sample mean is normal. If in the population distribution   the mean and variance value is respectively  (µ, σ), then the mean and variance for the distribution of the sample mean is(µ, σ/`sqrt(n)` ).
 For linearcombination the sample mean x=(1/n)(x₁+x₂+..x _n)
By using the above all concept the following theorem is illustrated.

Central Limit Theorem- distribution of the sample mean:
Theorem definition:

If we take a  population with a  mean μ and a variance σ², then the  sampling distribution of the mean approach a normal distribution with a mean of μ and a variance of σ²/N as N means   increases value of the sample size.

Irregular Triangle

A closed figure contains the three line of the segments that join end to end. The triangle is the three side polygon. The irregular triangle is the scalene triangle. Scalene triangles are abnormal in that they are distinct by what they are not. The majority triangles drained at chance would be scalene. The angles of the interior of the scalene triangle are normally all different.

Irregular triangle:

In the triangle all the sides are dissimilar length. The sides are not equal and angles are not equal.   The triangle is the general shape of geometry. This is containing the line segment. In the Euclidean geometry three non collinear positions is the represented to the unique triangle and unique or the unique plane. The right triangle is the other name for the right angled triangle. Hypotenuse triangle is the larger size of the right triangle. The length of the square is the equal to the length of the square of the hypotenuse.
Area of the irregular triangle:
The area of a triangle is usually computed as
               area = bh /2
Where b means base length and h means base height.
Calculating the length the side using the formula:
                 S= (a +b +c)/2

Irregular triangle fact:
Shortest side is opposite to the small angle:

• The straight side is constantly differing the minimum interior angle.
• The greatest side is constantly opposite the maximum interior angle.

Longest side is opposite to the largest angle:

• The straight side is constantly differing the nominal interior angle.
• The top side is forever differing the main interior angle.

Example problem:

Example 1:

Triangle EFG shown below is inscribed inside a square of side 40 cm. Find the area of the triangle.

Solution:
Area of triangle = (1/2) base *
= (1/2)(40)(40)
= 800 cm ²
Problem 
Calculating the third side from the specified right scalene triangle

Solution:
GI²=GH²+HI²
150=6²+HI²
150-36=HI²
114=HI²
HI² = `sqrt(114)`
HI = 10.67

Problem 3:
Fine the area of scalene triangle the problem, the given data :  s=6, a=5, b=3, c=4
Solution:
Area formula = `sqrt(s(s-a)(s-b)(s-c))`
 = sqrt(6(6-5)(6-3)(6-4))`
=sqrt(6(1)(3)(2))`
=sqrt(36)`
= 6
The area is 6cm.

Pitch Diameter Formula

When cutting screw threads the mechanic  must have some way to check the threads. Threads should be measured at the pitch diameter. The pitch diameter is a point along the flank or angular surface of the thread

. The thread pitch micrometer is an excellent and easy way to measure threads, but one pitch micrometer will not measure all of the different thread types and sizes. The thread ring gage will tell you whether the thread is right or wrong, but it will not give you an accurate account of the thread depth. The three-wire method is considered to be one of the most accurate and versatile ways of measuring threads. The three-wire method uses three lapped and polished wires and a micrometer to measure the pitch diameter of the thread

The wires are placed in the threads and a micrometer is used to measure over the top of the wires. Different thread types and thread sizes require different size wires. The three-wire method of thread measurement can be used for all types of threads including Unified, Acme, and Buttress thread forms.

To help you understand the three-wire method of thread measurement, we will use an example thread of 1.250-5 ACME-2G . The first piece of information we need to acquire is the "best wire size." The best wire size is the size of the wire that will contact the thread at the pitch diameter. The best wire size can be calculated using a formula, Begin by looking in the index under Thread, Acme, wire method of testing. Turn to this section and find the table which gives the wire sizes for measuring Acme threads with lead angles less than 5 degrees. Now look for the column that shows 5 threads per inch. You should find that the table gives you 0.10329 as the best wire size.  The table also gives you minimum and maximum wire size limits. Make a note of the best wire size.

Now we need to find out what the measurement over the wires should be. Within the same unit you will find a paragraph entitled Three Wire Measurement of Acme and Stub Acme Thread Pitch Diameters. Within this paragraph it will tell you where to find the Approximate Three Wire Formula. The Approximate Three Wire Formula "SIMPLIFIED VERSION" should read as follows:

M = E - 0.86603P + 3W

Where:

     M = the measurement over the wires

     E = the pitch diameter

     P = the pitch or (1 divided by the number of threads per inch)

    W= the wire size

We really only need to find the pitch diameter to solve the formula. We have the wire size. We know the pitch is 1/5 or 0.2 . To find the pitch diameter, go back to the index in the Machinery’s Handbook. Look in the index for Screw threads and threads systems. Under Screw threads and threads systems find Acme. Turn to this section and find the table that gives the Limiting Dimension for American National Standards for General Purpose Acme Single-Start Screw threads. In this table find the Class 2G, Pitch Diameter, Maximum and Minimum limiting diameters for the 1.250-5 ACME-2G thread. You should find that the table gives you 1.1210 as the minimum size and 1.1410 as the maximum size for the pitch diameter calculation. We will use the median dimension of 1.1310 for our formula calculation. Now that we have all of the information, let's plug the numbers into the formula.

M = E - 0.86603P + 3W

M = 1.131-(0.86603 X 0.2) + (3 X 0.10329)

M = 1.131 - 0.173206 + 0.30987

M = 1.268 (micrometer measurement over the wires)

Isosceles Triangle Proof

In an isosceles triangle, two sides are equal in length. An isosceles triangle also has two angles of the same measure; namely, the angles opposite to the two sides of the same length; this fact is the content of the Isosceles triangle theorem. Some mathematicians define isosceles triangles to have only two equal sides, whereas others define that an isosceles triangle is one with at least two equal sides. The latter definition would make all equilateral triangles isosceles triangles. (Source : WIKIPEDIA)

Things need to remember for Proofs of isosceles triangle:

For proving isosceles triangle we need to know the following information about  isosceles triangle,
There are two types of isosceles triangles,
1.Normal isosceles triangle
2.Right isosceles triangle.

Properties:
Properties of parts of  isosceles triangle:
1.The two sides of the isosceles tringles are equal.
2.Two base angles has same measure.
3. Angle ratio of the right isosceless triangle is   45:90:45.
4.The side ratio of the isosceles triangle is  1:1:`sqrt(2)`

Problems on isosceles triangle proof:

Problem 1:
Prove that the following triangle is isosceles triangle.

Proof:
Given , The angle We know that the sum of the angles are 180.
So 100 + x +x = 180
      100 + 2x = 180
Subtract 100 0n both sides.
    2x =180 -100
    2x = 80
Divide by x on both sides,
 x = 40
The base anles are equal two 40 .
According to the properties of isosceles triangle,we can determine that the given triangle is isosceles triangle.
Hence the proof.

Problem 2:
Prove that the triangle with the sides 5 : 5 : 5`sqrt(2)` is an isosceles right triangle triangle.
Proof:
Given,The sides of the triangle is 5 , 5 ,5`sqrt(2)`
We know that in a right Angle triangle,the hypotenuse is greater then the legs and it satisfies the pythagorean theorem,
5`sqrt(2)` > 5 , 5
5²+5² = (5`sqrt(2)` )²
25+25 = 25 *2
50 = 50
So the given sides satisfies the pythagoren theorem.So we can say that it is a right triangle.
The given sides are in the ratio of 1:1:`sqrt(2)`
so the given triangle is isosceles triangle.
Hence the proof.

Pyramid with a Trapezoid Base

A pyramid is a building where the outer surfaces are triangular and converge at a point. The base of a pyramid can be trilateral, quadrilateral, or any polygon shape, meaning that a pyramid has at least three outer surfaces (at least four faces including the base). The square pyramid, with square base and four triangular outer surfaces, is a common version. (Source: Wikipedia)

I like to share this Trapezoid Shape with you all through my article.

Pyramid with trapezoid base:
      A pyramid with a trapezoid base is called as trapezoidal pyramid.
     Volume of trapezoidal pyramid = (1 / 3) * ((1 / 2) * (base₁ + base₂) * height

Example problems for pyramid with a trapezoid base

Trapezoid pyramid problem 1:
        A base length of the trapezoid pyramid is 12 cm, 14 cm and its height of the pyramid is 6 cm. Find the volume of the trapezoidal pyramid.
Solution:
     Given base lengths are b₁ = 12 cm, b₂ = 14 cm, and h = 6 cm.
Formula:
   Volume of trapezoidal pyramid = (1 / 3) * ((1 / 2) * (base₁ + base₂) * height
Substitute the given values in the above formula, we get
= (1 / 3) * ((1 / 2) * (12 + 14) * 6 cm³
= (1 / 3) * 78 cm³
= 26 cm³

Answer:
 Volume of the trapezoidal pyramid is 26 cm³

Trapezoid pyramid problem 2:
        A base length of the trapezoid pyramid is 15 cm, 24 cm and its height of the pyramid is 9 cm. Find the volume of the trapezoidal pyramid.
Solution:
     Given base lengths are b₁ = 15 cm, b₂ = 24 cm, and h = 9 cm.
Formula:
Volume of trapezoidal pyramid = (1 / 3) * ((1 / 2) * (base₁ + base₂) * height
 Substitute the given values in the above formula, we get
= (1 / 3) * ((1 / 2) * (15 + 24) * 9 cm³
 = (1 / 3) * 175.5 cm³
                                               = 58.5 cm³
Answer:
 Volume of the trapezoidal pyramid is 58.5 cm³

Trapezoid pyramid problem 3:
        A base length of the trapezoid pyramid is 6 cm, 11 cm and its height of the pyramid is 5 cm. Find the area of the trapezoid.
Solution:
     Given base lengths are b₁ = 6 cm, b₂ = 11 cm, and h = 5 cm.
Formula:
Area of trapezoid = ((1 / 2) * (base₁ + base₂) * height
Substitute the given values in the above formula, we get
                                               = (1 / 2) * (6 + 11) * 5 cm³
                                               = (1 / 2) * 85 cm³
                                               = 42.5 cm²
Answer:
 Volume of the trapezoidal pyramid is 42.5 cm²

Practice problems for pyramid with a trapezoid base

Trapezoid pyramid problem 1:
        A base length of the trapezoid pyramid is 7 cm, 9 cm and its height of the pyramid is 12 cm. Find the volume of the trapezoidal pyramid.
Answer:
 Volume of the trapezoid pyramid is 32 cm³
Trapezoid pyramid problem 2:
        A base length of the trapezoid pyramid is 10 cm, 22 cm and its height of the pyramid is 7 cm. Find the volume of the trapezoidal pyramid.
Answer:

 Volume of the trapezoid pyramid is 37.33 cm³